Answer
$$
a(t)=v^{\prime}(t)=\sin t+3 \cos t \quad s(0)=0 , \quad v(0)=2
$$
The position of the particle is
$$
s(t)= -\sin t-3 \cos t +3t +3
$$
Work Step by Step
$$
a(t)=v^{\prime}(t)=\sin t+3 \cos t \quad s(0)=0 , \quad v(0)=2
$$
The general anti-derivative of $
v^{\prime}(t)=\sin t+3 \cos t
$ is
$$
v(t)= -\cos t+3 \sin t +C
$$
To determine C we use the fact that $v(0)=2$:
$$
v(0)= -\cos (0)+3 \sin (0) +C =2
$$
$ \Rightarrow $
$$
-1+C=2 \quad \Rightarrow \quad C=3,
$$
so
$$
v(t)= -\cos t+3 \sin t +3
$$
Since $s^{\prime }(t)=v(t)$, we antidifferentiate again and obtain
$$
s(t)= -\sin t-3 \cos t +3t +D
$$
To determine D we use the fact that $s(0)=0$:
$$
s(0)=-\sin (0)-3 \cos (0) +3(0)+D =0
$$
$ \Rightarrow $
$$
-3+D=0 \quad \Rightarrow \quad D=3,
$$
so the position of the particle is
$$
s(t)= -\sin t-3 \cos t +3t +3
$$
